Given an array arr[] of size 3 denoting the number of balls of type 1, 2, and 3 respectively, the task is to find the maximum number of moves that can be performed if in one move, three balls, out of which at least 2 balls of different types, are removed.
Examples:
Input: arr[] = {2, 3, 3}
Output: 2
Explanation:
Move 1: Remove 1 ball of each type. Therefore, arr[] becomes {1, 2, 2}.
Move 2: Remove 1 ball of each type. Therefore, arr[] becomes {0, 1, 1}.
No further moves can be performed.Input: arr[] = {100, 1, 2}
Output: 3
Explanation:
Move 1: Remove 1 ball of type 2 and 2 balls of type 1. Therefore, arr[] becomes {98, 0, 2}.
Move 2: Remove 1 ball of type 3 and 2 balls of type 1. Therefore, arr[] becomes {96, 0, 1}.
Move 3: Remove 1 ball of type 3 and 2 balls of type 1. Therefore, arr[] becomes {94, 0, 0}.
No further moves can be performed.
Approach: The idea to sort the array in increasing order and then arises two cases:
- If arr[2] is smaller than 2 * (arr[0] + arr[1]), the answer will be (arr[0] + arr[1] + arr[2]) / 3.
- If arr[2] is greater than 2 * (arr[0] + arr[1]), the answer will be arr[0] + arr[1].
Follow the steps below to solve the problem:
- Sort the array in increasing order.
- Print the minimum of (arr[0]+arr[1]+arr[2])/3 and arr[0]+arr[1].
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to find maximum moves that // can be performed if in each move 3 // balls of different types are removed void findMoves( int arr[])
{ // Sort the array in increasing order
sort(arr, arr + 3);
// Print the answer
cout << (min(arr[0] + arr[1],
(arr[0] + arr[1] + arr[2]) / 3));
} // Driver Code int main()
{ // Given Input
int arr[3] = { 2, 3, 3 };
// Function Call
findMoves(arr);
return 0;
} |
// Java program for the above approach import java.util.Arrays;
class GFG{
// Function to find maximum moves that // can be performed if in each move 3 // balls of different types are removed static void findMoves( int arr[])
{ // Sort the array in increasing order
Arrays.sort(arr);
// Print the answer
System.out.println(Math.min(arr[ 0 ] + arr[ 1 ],
(arr[ 0 ] + arr[ 1 ] +
arr[ 2 ]) / 3 ));
} // Driver Code public static void main(String[] args)
{ // Given Input
int arr[] = { 2 , 3 , 3 };
// Function Call
findMoves(arr);
} } // This code is contributed by AnkThon |
# Python3 program for the above approach # Function to find maximum moves that # can be performed if in each move 3 # balls of different types are removed def findMoves(arr):
# Sort the array in increasing order
arr = sorted (arr)
# Print the answer
print ( min (arr[ 0 ] + arr[ 1 ],
(arr[ 0 ] + arr[ 1 ] +
arr[ 2 ]) / / 3 ))
# Driver Code if __name__ = = '__main__' :
# Given Input
arr = [ 2 , 3 , 3 ]
# Function Call
findMoves(arr)
# This code is contributed by mohit kumar 29 |
// Java program for the above approach using System;
class GFG
{ // Function to find maximum moves that // can be performed if in each move 3 // balls of different types are removed static void findMoves( int []arr)
{ // Sort the array in increasing order
Array.Sort(arr);
// Print the answer
Console.Write(Math.Min(arr[0] + arr[1],
(arr[0] + arr[1] +
arr[2]) / 3));
} // Driver Code public static void Main(String[] args)
{ // Given Input
int []arr = { 2, 3, 3 };
// Function Call
findMoves(arr);
} } // This code is contributed by shivanisinghss2110 |
<script> // JavaScript program for the above approach // Function to find maximum moves that
// can be performed if in each move 3
// balls of different types are removed
function findMoves(arr) {
// Sort the array in increasing order
arr.sort( function (a, b) { return a - b; })
// Print the answer
document.write(Math.min(arr[0] + arr[1],
parseInt((arr[0] + arr[1] + arr[2]) / 3)));
}
// Driver Code
// Given Input
let arr = [2, 3, 3];
// Function Call
findMoves(arr);
// This code is contributed by Potta Lokesh
</script>
|
2
Time Complexity: O(1)
Auxiliary Space: O(1)